## Calculus 8th Edition

$\dfrac{1}{3}$
$\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}$ Factor: $\lim\limits_{t \to 2}\dfrac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$ Simplify: $\lim\limits_{t \to 2}\dfrac{t+2}{t^2+2t+4}$ Plug in 2: $\dfrac{2+2}{2^2+2(2)+4}=\dfrac{4}{12}=\dfrac{1}{3}$ Therefore, $\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}=\dfrac{1}{3}$